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I just want to know how this circuit functions.  All I know is that a diode allows current in one direction only and the capacitors store a charge, resistors dissipate some power and the LED is a power indicator.

 

So, how does this really work?  Okay, the capacitors store the charge.  Does the capacitor just store or does it also provide some output power?  And what about the in/out/adj, what is this for?  My plan is to charge my 6 volt battery.

Any help would be appreciated.

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First off, I would not recommend using this voltage regulator to charge a battery.  A few things could happen:

  • The battery could overheat because of over-current
  • There is no circuitry to tell it to stop charging once the battery is at full charge

Now this is not to say you could not charge a battery with it.  I’m just saying there are things that could go wrong and cause problems.

So let’s see what your circuit does.  Let’s start with your regulator which has “in”, “out” and “adj”.

The datasheet on the LM317 says that it is a 1.2 volt regulator.  What this means is that if you place ground on the “adj” pin then you will have an output of 1.2 volts as long as the input is adequate for providing this.  Your circuit says, over on the far right, that there is a 7.2 volt output.  In order to obtain the stated output we have to change the voltage on the “adj” pin to your desired output voltage minus 1.2 volts.

So we need 6 volts on the “adj” pin to get our desired output.  So R1 and R2 are selected to build a voltage divider that will provide 6 volts for the “adj” pin.  Here is how we do that.

So you see the resistors selected provide the needed 6 volts reference at the “adj” pin.  Now let’s address the first part of your question, “Does the capacitor just store or does it also provide some output power?”  If you think about it, there is no reason for the capacitor to store voltage unless we are going to need it.  Your circuit has three capacitors.  They all store a voltage and sit there doing nothing as long as that voltage does not change.  If that voltage changes, up or down, the capacitor has to react to it.  If the voltage goes up then the capacitor charges more.  If the voltage goes down then the capacitor discharges it stored charge, back into the circuit.  The capacitors job is to minimize these ups and downs.

Your next question, “And what about the in/out/adj, what is this for?”  We already discussed the “out” and the “adj”, so what happens at the “in”?  At the “in” we have to have enough voltage to provide the output voltage plus any losses that may take place in the regulator.  Every regulator may be a little different but, for our circuit, the “in” has to be adequately above our “out” of 7.2 volts and the 12 volts on the anode of diode D1 accomplishes that.

Now let’s discuss what diodes D1, D2 and D3 do for the circuit.  Their only purpose is for protection.  If things go wrong, they have to reroute current to protect your load and your regulator.

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